The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) - - > 4CO2 (g) + 2H2O(g) How many liters of CO 2 are produced when 52.0 g C2H 2 burns in oxygen? (Assume STP.)

1 Answer
Nam D.
Mar 5, 2018

I get #90.8 \ "L"#.

Explanation:

We have the balanced equation:

#2C_2H_2(g)+5O_2(g)->4CO_2(g)+2H_2O(g)#

At #STP#, the molar volume is #22.7 \ "L"#.

First, we must convert #52# grams of acetylene into moles. Acetylene has a molar mass of #26.04 \ "g/mol"#. So here, there exist approximately

#(52color(red)cancelcolor(black)"g")/(26.04color(red)cancelcolor(black)"g""/mol")~~2 \ "mol"#

Now, we see that the mole ratio between #C_2H_2# is #2:4#. So, #2# moles of acetylene will fully produce #4# moles of #CO_2#.

Since we also have #2# moles of acetylene, we produce #4# moles of #CO_2#. Remember at #STP#, the molar volume is #22.7 \ "L"#, and so the volume of carbon dioxide produce would be

#4color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(1color(red)cancelcolor(black)"mol")=90.8 \ "L"#

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